%I #37 Nov 01 2019 12:16:58

%S 1,10,41,174,737,3122,13225,56022,237313,1005274,4258409,18038910,

%T 76414049,323695106,1371194473,5808472998,24605086465,104228818858,

%U 441520361897,1870310266446,7922761427681,33561355977170,142168185336361,602234097322614

%N Generalized Pellian with second term of 10.

%H Reinhard Zumkeller, <a href="/A048879/b048879.txt">Table of n, a(n) for n = 0..1000</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,1)

%F a(n) = ((8+sqrt(5))*(2+sqrt(5))^n - (8-sqrt(5))*(2-sqrt(5))^n)2*sqrt(5).

%F From _Philippe Deléham_, Nov 03 2008: (Start)

%F a(n) = 4*a(n-1) + a(n-2); a(0)=1, a(1)=10.

%F G.f.: (1+6*x)/(1-4*x-x^2). (End)

%F For n >= 1, a(n) equals the denominator of the continued fraction [4, 4, ..., 4, 10] (with n copies of 4). The numerator of that continued fraction is a(n+1). - _ZhenShu Luan_, Aug 05 2019

%p with(combinat): a:=n->6*fibonacci(n-1,4)+fibonacci(n,4): seq(a(n), n=1..16); # _Zerinvary Lajos_, Apr 04 2008

%t LinearRecurrence[{4,1},{1,10},30] (* _Harvey P. Dale_, Jul 18 2011 *)

%o (Haskell)

%o a048879 n = a048879_list !! n

%o a048879_list = 1 : 10 : zipWith (+)

%o a048879_list (map (* 4) $ tail a048879_list)

%o -- _Reinhard Zumkeller_, Mar 03 2014

%Y Cf. A015448, A001077, A001076, A033887.

%K easy,nice,nonn

%O 0,2

%A _Barry E. Williams_

%E More terms from _Harvey P. Dale_, Jul 18 2011